The Generalized Newtonian Fluid - Isothermal Flows Constitutive Equations! Viscosity Models! Solution of Flow Problems!

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1 The Geeralized Newtoia Fluid - Isothermal Flows Costitutive Equatios! Viscosity Models! Solutio of Flow Problems! 0.53/2.34! Sprig 204! MIT! Cambridge, MA 0239!

2 Geeralized Newtoia Fluid Simple Shear Flow Arbitrary Flow Newtoia Fluid Newtoia Fluid y x dvx τ yx = µ dy dvj τ ij = + µ dx i = µγ ij dv i dx j No-Newtoia Fluid y where x dv τ η x yx = dy η = η( dv ) x dy No-Newtoia Fluid where τ ij = ηγ η = ηγ ( ) γ = 2 II ij γ 2

3 No-Newtoia Viscosity η ( ) I γ = i ii γ = 2 v = 0 II γ = γ γ i j ij ji = 2 γ III γ = γ ijγ jk ki γ = 0 i j j defiitio of γ = 2 II γ shear rate 3

4 Geeralized Newtoia Fluid shear stressessteady shear flow ( γ ) η 4

5 Power-Law Model η = mγ η 0 slope = - Captures high shear rate behavior ormally occurig i processig log η m η < = > shear thiig [0.5 < < 0.6] Newtoia fluid m = µ shear thickeig log γ Defects No η 0 No λ 5

6 Spriggs "Trucated Power-Law" η = η 0 γ γ 0 γ η = η 0 γ 0 γ > γ 0 Cotais a characteristic time λ λ = γ 0 log η η 0 γ 0 slope = - η log γ 02 6

7 Carreau-Yasuda Model! $!%! $! 0 % [ ] = +( "# ) a $ a! (Pa s) "# (s - ) 7

8 Bigham Model η = τ γ ( τ τ ) 0 0 µ + ( τ τ ) 0 0 Bigham plastics have a yield stress τ 0 ad will ot flow uless the magitude of the stress τ exceeds τ 0 - τ yx = ηγ τ0 slope µ 0 τ = 2 ( : ) γ Time costat µ 0 τ 0 8

9 Casso Model dvx ± τyx = τ0 + µ 0 m for τ yx > τ dy γ yx = 0 for τyx < τ 0 Useful for chocolate 0 9

10 Tube Flow of a Power-Law Fluid ( ) p 0 p L πr 2 τ r z 2πrL= 0 R r z cotrol volume for mometum balace r τrz = τ w R pr τ w = 2L p 0 L p L dv τ τ = rz η( γ) γ rz = mγ dr τ rz dvz = m dr z 0

11 Tube Flow Results v z =!, # " m $ R & r ( %! # " + ( R$ ' + w Q = 3 - R &, w + 3 '( m. R =! 3 + #. " 4 $ ) *+ ) + + * true wall shear rate a = 8 v. a z D apparet (Newtoia) wall shear rate

12 Radial Flow betwee Parallel Problem: Disks Determie Q(p -p 2, B, R, R 2, m, ) Assume the fluid viscosity is described by the power-law fuctio Solutio: Use the lubricatio approximatio to simplify the problem to pressure drive slit flow First fid Q for this simple flow 2

13 Assume v v ( x ), v v 0 z z x y ij ij () x z-compoet of the equatio of motio (Table B.) v v v v p v v v g t x y z x y z z z z z z x y z xz yz zz z d dx xz p L p 0 L Itegrate to get the shear stress distributio from coservatio of mometum xz L p0 pl x c 0 3

14 For the geeralized Newtoia fluid xz dvz dx ad the power-law model gives dvz m m dx I order to avoid problems with the absolute value, cosider oly the regio x > 0, for which dvz dvz dvz 0 dx dx dx The xz dvz dvz dvz x m m ( p0 pl ) dx dx dx L 4

15 The differetial equatio for the velocity is thus dv p p dx ml z 0 L which ca be itegrated to give Fially, the volume flow rate is foud as p 0 pl vz B x ml x where B 2 2WB B Q 2W vz dx 2 m B 0 p0 pl dp B B L dz 5

16 Apply the slit flow results locally to the disk problem The correspodig quatities i the two geometries are pressure gradiet width p0 pl dp dp L dz dr W 2r The volume flow rate expressio is adapted as 2 Q m dp 2 2B 2r B dr which ca be itegrated from r = R to r = R 2, by takig advatage of the fact that for icompressible fluids Q is idepedet of r. This gives 6

17 p Q m 2 R R p 4 B B ( ) Fially, the above is iverted to give the volume flow rate i terms of the pressure gradiet Q 2 4B 2 p p B( ) m R 2 R 7

18 Justificatio for Applyig the Lubricatio Approximatio Use a order of magitude aalyis to justify the use of the lubricatio approximatio i adaptig the slit flow results to the radial disk flow problem For the radial flow problem, assume that v v ( r,z ), v v 0 r r z r-compoet of the equatio of motio v r p vr rrr zr r r r z r r For the power-law fluid 8

19 For the power-law fluid m ; : ij ij 2 From Table B.3 i DPL, we get the rate-of-strai tesor i cylidrical coordiates vr vr 2 0 r z vr r v r 0 0 z from which the shear rate is foud vr vr vr 2 2 r r z 9

20 Order of magitude estimates for cotributios to the shear rate v for R << R 2 r Q V 4 RB vr V vr V vr V ; ; r R r R z B 2 2 For small gap, B << R 2, the shear rate is well approximated by v r z V B Next evaluate the order of magitude of the terms that appear i the equatio of motio (Iertial term) v r v r r V R

21 (Stress terms) ; ; 2 zr mv mv m V V rrr r r R B z B B r R B R Compariso of differet terms shows that zr is the largest term o the right side by a factor of BR 2 2 z The ratio of iertial to viscous forces is 2 V R2 VR 2 B B Re m V mvb R 2 R 2 B B 2 2 If the Reyolds umber is at most (R 2 /B), the iertial forces ca still be eglected 2

22 Neglectig terms of order (B/R 2 ) ad smaller gives the equatio of motio as p r z 0 zr which is locally (i r) the same as the slit flow equatio 22

24 Equate this to the power-law result for Q(dp/dz) for tube flow Q 0 3 z R R dp L 2m dz 3 Itegrate to get the pressure drop dow the tube p 3 z 3 R R L 2mL 0 p a Q At ay positio z, there is p - p a drivig force to force fluid through the slit of local legth l(z) Slit flow for a power law fluid gives V B 2 p pa B 2 2 ml( z) 24

25 Equatig the available to eeded pressure gradiet gives l(z) BL B 3 Q0 z () 3 lz R ( ) 2R 2 V L 25

26 Squeezig Flow betwee Parallel Disks (DPL Example 4.2-7) 26

27 Volume flow rate across surface at r Mass coservatio 2 2 Qr r h Equatio of motio with the lubricatio approximatio Slit flow W B 0 L P P L Q Radial flow 2r h dp dr Qr Hece Qr 2 rh h dp 22 2 m dr 27

28 Solve for p(r) with the boudary coditio that p(r) = p a p m h 2 R r pa 2 h 2 R A force balace o the top plate gives 2 R Ft () p p rdrd 0 0 a zz zh = 0 o ay solid surface F h 2 mr 2 h Scott equatio 28

29 For costat force this ca be itegrated to give the half time t /2 for h to go from h 0 to (½)h 0 : 2 t2 Rm R K F h0 fuctio of 29

30 where m 2m m Viscous respose = De - VRXUFHXQNQRZQ$OOULJKWVUHVHUYHG7KLVFRQWHQWLVH[FOXGHGIURPRXU&UHDWLYH &RPPRQVOLFHQVH)RUPRUHLQIRUPDWLRQVHHKWWSVRFZPLWHGXKHOSIDTIDLUXVH 30

31 MIT OpeCourseWare J / 0.53J Macromolecular Hydrodyamics Sprig 206 For iformatio about citig these materials or our Terms of Use, visit:

Fluid Physics 8.292J/12.330J % (1)

Fluid Physics 8.292J/12.330J % (1) Fluid Physics 89J/133J Problem Set 5 Solutios 1 Cosider the flow of a Euler fluid i the x directio give by for y > d U = U y 1 d for y d U + y 1 d for y < This flow does ot vary i x or i z Determie the